B Integral with Exponential Integral Function

[EngWiPy]

Hi all,

I have this integral:

\int_0^{\infty}\nu^{n}e^{-j\nu[y+n]}\left[E_1(-j\nu)\right]^n\,d\nu

How can I evaluate this integral? I found the attached integral formula in the table of integrals, but the exponential integral function $E_1(.)$ is not raised to the power of an exponent $n\ge 0$. Any idea?

Thanks in advance

 

[Ssnow]

Hi, to find an exact answer seems hard! especially for every $n$. I suggest you to try with a math software as Mathematica or use some approximation of $E_{1}(x)$ as the following:

$E_{1}(x)=\int_{x}^{+\infty}\frac{e^{-t}}{t}dt \sim e^{-x}\sum_{n=0}^{+\infty}(-1)^{n}\frac{n!}{x^{n+1}}$

that is a good approximation for $x\rightarrow +\infty$.

 

[EngWiPy]

Hi, to find an exact answer seems hard! especially for every $n$. I suggest you to try with a math software as Mathematica or use some approximation of $E_{1}(x)$ as the following:

$E_{1}(x)=\int_{x}^{+\infty}\frac{e^{-t}}{t}dt \sim e^{-x}\sum_{n=0}^{+\infty}(-1)^{n}\frac{n!}{x^{n+1}}$

that is a good approximation for $x\rightarrow +\infty$.

I could use numerical integration I guess, but this isn't the final result in my analysis. The infinite series approximation is still problematic as it's raised to the power and its upper limit is infinite.

 

[zaidalyafey]

Well, I suggest you use integration by parts since

$$\frac {\partial }{\partial x} E(-xy)^n =-n\frac {e^{xy}}{x} E(-xy)^{n-1} $$

I think it is easy for small values of n. Try to generalize it.

 

[EngWiPy]

Well, I suggest you use integration by parts since

$$\frac {\partial }{\partial x} E(-xy)^n =-n\frac {e^{xy}}{x} E(-xy)^{n-1} $$

I think it is easy for small values of n. Try to generalize it.

Thanks. Do you have a resource for this formula?

 

[zaidalyafey]

Thanks. Do you have a resource for this formula?

It is a simple differentiation of power

$$\frac {d}{dx}[g (x)]^n = n g'(x) [g (x)]^{n-1}$$

 

[EngWiPy]

It is a simple differentiation of power

$$\frac {d}{dx}[g (x)]^n = n g'(x) [g (x)]^{n-1}$$

Right. But how to differentiate $E_1(-j\nu)$. I know that

E_1(-j\nu)=\int_1^{\infty}\frac{e^{j\nu t}}{t}\,dt.

This means that
\frac{\partial}{\partial \nu}E_1(-j\nu)=\int_1^{\infty}\frac{\partial}{\partial \nu} \frac{e^{j\nu t}}{t}\,dt=j\int_1^{\infty}e^{j\nu t}\,dt

Right? How did you get your expression?

Thanks

 

[EngWiPy]

Am I right in this or not?

 

[Ssnow]

hhmmm, it is convenient to have the integral in the form $E_{1}(-j\nu)=\int_{-j\nu}^{+\infty}\frac{e^{-t}}{t}dt$ so you can apply the fundament theorem of calculus when you derive respect $\nu$...
To be precise it is better take the limit:

$\lim_{b\rightarrow +\infty}\frac{\partial}{\partial\nu}\int_{-j\nu}^{b}\frac{e^{-t}}{t}dt =...$

 

[zaidalyafey]

Yes as Ssnow suggested. Can you try sovle the integral for n=2 ?

Sorry I for the late response I was travelling.